c) 3^{x + 4} + 3^{x + 2} = 810

=> 3^x . 3⁴ + 3^x . 3² = 810

=> 3^x.(3⁴ + 3²)       = 810

=> 3^x . (81 + 9)      = 810

=> 3^x . 90               = 810

=> 3^x                      = 810 : 90

=> 3^x                      = 9

=> 3^x                      = 3²

=>    x                     = 2

d) 3^x + 3^{x + 2} = 810

=> 3^x + 3^x . 3² = 810

=> 3^x . (1 + 3²) = 810

=> 3^x . (1 + 9)   = 810

=> 3^x . 10          = 810

=> 3^x                 = 810 : 10

=> 3^x                  = 81

=> 3^x                  = 3⁴

=>   x                  = 4

c, \(3^{x+4}+3^{x+2}=810\)

\(\Leftrightarrow3^x\left(3^4+3^2\right)=810\)

\(\Leftrightarrow3^x.90=810\)

\(\Leftrightarrow3^x=9=3^2\)

\(\Leftrightarrow x=2\)

d, \(3^x+3^{x+2}=810\)

\(\Leftrightarrow3^x\left(1+3^2\right)=810\)

\(\Leftrightarrow3^x.10=810\)

\(\Leftrightarrow3^x=81=3^4\)

\(\Leftrightarrow x=4\)

P/s: Toán thường thôi nhỉ :) Ko nâng cao lắm

#)Giải :

c) 3^x+4 + 3^x+2 = 810

=> 3^x.3⁴ + 3^x.3² = 810

=> 3^x(3⁴ + 3²) = 810

=> 3^x(81 + 9) = 810

=> 3^x.90 = 810

=> 3^x = 9 = 3²

=> x = 2 

d) 3^x + 3^x+2 = 810

=> 3^x.1 + 3^x.3² = 810

=> 3^x(1 + 3²) = 810

=> 3^x.10 = 810

=> 3^x = 81 = 3⁴ 

=> x = 4

hình như sai đề

sai thật mày ạ

3^x + 3^x+2 = 810

3^x + 3^x.3² = 810

3^x(1+9) = 810

3^x.10 = 810

3^x = 810 : 10

3^x = 81 = 3⁴

=> x = 4

3^x+3 + 3^x+2 = 810

=> 3^x ( 1 + 3² ) = 810

=> 3^x . 10 = 810

=> 3^x = 81

=> 3^x = 3⁴

=> x = 4

`#3107.101107`

a)

\(27< 3^x< 243\\ \Rightarrow3^3< 3^x< 3^5\\ \Rightarrow3< x< 5\\ \Rightarrow x=4\)

Vậy, `x = 4`

b)

\(2^x+2^{x+1}+2^{x+2}=56?\\ \Rightarrow2^x+2^x\cdot2+2^x\cdot4=56\\ \Rightarrow2^x\cdot\left(1+2+4\right)=56\\ \Rightarrow2^x\cdot7=56\\ \Rightarrow2^x=8\\ \Rightarrow2^x=2^3\\ \Rightarrow x=3\)

Vậy, `x = 3`

c)

\(3^x+3^{x+2}=810\\ \Rightarrow3^x+3^x\cdot9=810\\ \Rightarrow3^x\cdot\left(1+9\right)=810\\ \Rightarrow3^x\cdot10=810\\ \Rightarrow3^x=81\\ \Rightarrow3^x=3^4\\ \Rightarrow x=4\)

Vậy, `x = 4.`

a) \(27< 3^x< 243\)

\(\Rightarrow3^3< 3^x< 3^5\)

\(\Rightarrow3< x< 5\)

c) \(3^x+3^{x+2}=810\)

\(\Rightarrow3^x\left(1+3^2\right)=810\)

\(\Rightarrow3^x.10=810\)

\(\Rightarrow3^x=810:10\)

\(\Rightarrow3^x=81\)

\(\Rightarrow3^x=3^4\)

\(\Rightarrow x=4\)

a, Ta có : 

\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\Rightarrow\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x+3y-z-2-6+3}{4+9-4}=\dfrac{50-5}{9}=5\)

\(\Rightarrow x=11;y=17;z=23\)

b, Đặt \(\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\Rightarrow xyz=810\)

\(\Rightarrow2k.3k.5k=810\Leftrightarrow30k^3=810\Leftrightarrow k^3=27\Leftrightarrow k=3\)

\(\Rightarrow x=6;y=9;z=15\)

a) Ta có: \(\dfrac{x-1}{2}=\dfrac{2x-2}{4};\dfrac{y-2}{3}=\dfrac{3y-6}{9};\dfrac{z-3}{4}\)

Áp dụng t/c dtsbn:

\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x-2+3y-6-z+3}{4+9-4}=5\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=5\\\dfrac{y-2}{3}=5\\\dfrac{z-3}{4}=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=11\\y=17\\z=12\end{matrix}\right.\)

b) Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)

xyz = 810

=> 2k.3k.5k = 810

=> k = 3

\(\Rightarrow\left\{{}\begin{matrix}x=6\\y=9\\z=15\end{matrix}\right.\)

a) Ta có: \(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)

nên \(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)

mà 2x+3y-z=50

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x+3y-z-2-6+3}{4+9-4}=\dfrac{50-5}{9}=5\)

Do đó:

\(\left\{{}\begin{matrix}x-1=10\\y-2=15\\z-3=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=11\\y=17\\z=23\end{matrix}\right.\)

b) Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)

Ta có: xyz=810

\(\Leftrightarrow30k^3=810\)

\(\Leftrightarrow k^3=27\)

\(\Leftrightarrow k=3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=2\cdot3=6\\y=3k=3\cdot3=6\\z=5k=5\cdot3=15\end{matrix}\right.\)

\(3^x+3^{x+2}=810\)

=> \(3^x\cdot1+3^x\cdot3^2=810\)

=> \(3^x\cdot\left(1+3^2\right)=810\)

=> \(3^x\cdot10=810\)

=> \(3^x=810:10=81\)

=> \(x=4\)

3^x+2+3^x=810

=> 3^x.(3²+1)=810

=> 3^x.(9+1)=810

=> 3^x.10=810

=> 3^x=810:10

=> 3^x=81

=> 3^x=3⁴

=> x=4

3^x+2+3^x=810

=> 3^x.(3²+1)=810

=> 3^x.(9+1)=810

=> 3^x.10=810

=> 3^x=810:10

=> 3^x=81

=> 3^x=3⁴

=> x=4

chúa ạ mấy câu dễ   

x=6;y=9;z=15 à